\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

\(3x-2x^2=0\)

\(x\left(2x-3\right)=0\)

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

Bài 1

1.(x-3)(x+2)-x(x-7)=15

\(\Leftrightarrow x^2+2x-3x-6-x^2+7x=15\)

\(\Leftrightarrow-6+6x=15\)

\(\Leftrightarrow6x=15+6\) =21

\(\Rightarrow x=\dfrac{21}{6}=3,5\)

2.(x-5)(x+5)+x(3-x)=20

\(\Leftrightarrow x^2-25+3x-x^2=20\)

\(\Leftrightarrow-25+3x=20\)

\(\Leftrightarrow3x=20+25=45\)

\(\Rightarrow x=\dfrac{45}{3}=15\)

3.(x-7)²-x(2+x)=-7

\(\Leftrightarrow x^2-14x+49-2x-x^2=-7\)

\(\Leftrightarrow-16x+49=-7\)

\(\Leftrightarrow-16x=-7-49=-56\)

\(\Rightarrow x=\dfrac{-56}{-16}=\dfrac{7}{2}=3,5\)

Tiếp bài 1

4.(x-4)²-(x+4)(x-4)=-16

\(\Leftrightarrow x^2-8x+16-x^2-16=-16\)

\(\Leftrightarrow-8x=-16\)

\(\Rightarrow x=\dfrac{-16}{-8}=2\)

5.(x-5)(x+5)-x(2-3x)=4x²-7

\(\Leftrightarrow x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-25-2x+3x^2=4x^2-7\)

\(\Leftrightarrow4x^2-4x^2-2x=-7+25\)

\(\Leftrightarrow-2x=18\)

\(\Rightarrow x=\dfrac{18}{-2}=-9\)

B2

1. 2x(x-2)+3x-6=0

=>2x(x-2)+3(x-2)=0

=>(x-2)(2x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\2x+3=0\Leftrightarrow x=\dfrac{-3}{2}\end{matrix}\right.\)

2.3x(x-5)-5x+25=0

=>3x(x-5)-5(x-5)=0

=>(x-5)(3x-5)=0

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\3x-5=0\Leftrightarrow x=\dfrac{5}{3}\end{matrix}\right.\)

3. x(x+7)-x-7=0

=>x(x+7)-(x+7)=0

=>(x+7)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Leftrightarrow x=-7\\x-1=0\Leftrightarrow x=1\end{matrix}\right.\)

(Ko chép lại đề)

\(a.\Rightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

\(b.\left(3x-2\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=7\end{cases}}\)

\(c.7x^2=28\)

\(x^2=4\)

\(x^2=2^2\)

\(x=\pm2\)

\(d.\left(2x+1\right)\left(1+x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}\)

a)\(\left(3x+5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)

b)\(3x\left(x-7\right)-2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{2}{3}\end{cases}}}\)

c) \(7x^2-28=0\)

\(\Leftrightarrow7x^2=28\)

\(\Leftrightarrow7x^2=7.4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

d)\(\left(2x+1\right)+x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(1+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}}\)

#H

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

Bài 1.

a) x² + 7x +12 = 0

Ta có Δ = 7² - 4.12 = 1> 0 => \(\sqrt{\Delta}=\sqrt{1}=1\)

Phương trình có 2 nghiệm phân biệt:

x₁ = \(\frac{-7+1}{2}=-3\)

x₂= \(\frac{-7-1}{2}=-4\)

Bài 1

b) 2x² + 5x - 3=0

Ta có: Δ = 5² + 4.2.3 = 49 > 0 => \(\sqrt{\Delta}=\sqrt{49}=7\)

Phương tình có 2 nghiệm phân biệt:

x_{1 = \(\frac{-5+7}{2.2}=\frac{1}{2}\)}

x₂ = \(\frac{-5-7}{2.2}-3\)

c) 3x² +10x+7 = 0

Ta có: Δ = 10² - 4.3.7= 16> 0 => \(\sqrt{\Delta}=\sqrt{16}=4\)

Phương tình có 2 nghiệm phân biệt:

x₁= \(\frac{-10+4}{2.3}=-1\)

x₂= \(\frac{-10-4}{2.3}=-\frac{7}{3}\)

Bài 1

d)x4+5x2-36=0

Đặt x² = t ( đk: t ≥0)

=> t² +5t - 36 =0

Ta có: Δ = 5² + 4.36 = 169 > 0 => \(\sqrt{\Delta}=\sqrt{169}=13\)

Phương tình có 2 nghiệm phân biệt:

t₁ = \(\frac{-5-13}{2}=-9\) (loại)

t₂ = \(\frac{-5+13}{2}=4\) (thỏa mãn)

Với t = 4 ta có:

x² = 4

\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)